Resulting force of an object with variable movement

[Thietlan]

Homework Statement

The movement of a material point of mass 500 mg is described using the following formula: $v=9 \cdot \beta \cdot t^4$ ,variable in time and is expressed in km/h. Considering the movement in the context of classical physics, and knowing the constant $\beta=0.2 m \cdot s^{-5}$ . The resulting force which acts on the object at t=20s is equal to:
$0.08 kg m \cdot s^{-5}$
$8 kg *m \cdot s^{-2}$
80 N
8 kN

Relevant Equations

$v=9* \beta * t^4$
$\beta = 0.2 m*s^{-5}$

v(20)=288000 m/s=80000km/h
a=(80 000-0)/(20 -0)=4000 m*s^(-2)
F=m*a=0.0005*80 000 which doesn't fit any answers

 

[PeroK]

You seem to have assumed constant force and constant acceleration.

 

[Thietlan]

You seem to have assumed constant force and constant acceleration.

Then how should I go about solving it?

 

[PeroK]

Then how should I go about solving it?

Treat the acceleration as a function of time.

 

[Thietlan]

Treat the acceleration as a function of time.

Ok I calculate acceleration as the derivative of v which comes out to be $a=36 \cdot \beta t^3 km/h^2$ and then simply calculate F= m*a(20) ,still doesn't give the right answer. What else am I missing?

 

[PeroK]

Ok I calculate acceleration as the derivative of v which comes out to be $a=36 \cdot \beta t^3 km/h^2$ and then simply calculate F= m*a(20) ,still doesn't give the right answer. What else am I missing?

I must be honest, I don't understand the question. It gives $\beta$ in $m/s$, time in seconds and force in SI units. So, I don't understand the reference to $km/h$.

 

[Steve4Physics]

A confusingly written question. But you can match one of the answers in the list by proceeding as follows:

v is the velocity in km/h. Let V be the velocity in m/s. That means V = v/3.6 and the given equation becomes:
V(t) = 9βt⁴/3.6 = 9*0.2*t⁴/3.6 = 0.5t⁴

Differentiate V(t) to find a(t) in m/s². The rest is straightforward.

Edit: minor changes

 

[PeroK]

A confusingly written question. But you can match one of the answers in the list by proceeding as follows:

v is the velocity in km/h. Let V be the velocity in m/s. That means V = v/3.6 and the given equation becomes:
V(t) = 9βt⁴/3.6 = 9*0.2*t⁴/3.6 = 0.5t⁴

Differentiate V(t) to find a(t) in m/s². The rest is straightforward.

Edit: minor changes

You did well to work that out. The question, IMO, is the product of a sick mind!

 

[Thietlan]

I must be honest, I don't understand the question. It gives $\beta$ in $m/s$, time in seconds and force in SI units. So, I don't understand the reference to $km/h$.

Neither did I and yes while dividing by 3.6 gives one of the responses I don't really get why I would need to divide. If it is in km/h then we should be multiplied by 3.6 and if it were in m/s well,in case there is no need to convert.

 

[Steve4Physics]

Neither did I and yes while dividing by 3.6 gives one of the responses I don't really get why I would need to divide. If it is in km/h then we should be multiplied by 3.6 and if it were in m/s well,in case there is no need to convert.

For example, take v = 3.6km/h. What is this in m/s?

This speed corresponds to covering 3600m in 3600s. In m/s, this speed is V = distance/time = 3600m/3600s = 1m/s. (There are other ways to convert units, but this method should make what is happening clear.)

In terms of simple arithmetic we have divided the speed in km/h by 3.6 to get the speed in m/s.

In the given equation v(t) = 9βt⁴, we are (confusingly) told that v is in km/h. So dividing v by 3.6 converts it to m/s.

If V is the velocity in m/s: V(t) = v(t)/3.6 = 9βt⁴/3.6

And from that point on, using V(t), we are working entirely in SI units.